\(\int \frac {1}{(c \sin (a+b x))^{5/2}} \, dx\) [31]

Optimal result

Integrand size = 12, antiderivative size = 77 \[ \int \frac {1}{(c \sin (a+b x))^{5/2}} \, dx=-\frac {2 \cos (a+b x)}{3 b c (c \sin (a+b x))^{3/2}}+\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right ),2\right ) \sqrt {\sin (a+b x)}}{3 b c^2 \sqrt {c \sin (a+b x)}} \]

[Out]

-2/3*cos(b*x+a)/b/c/(c*sin(b*x+a))^(3/2)-2/3*(sin(1/2*a+1/4*Pi+1/2*b*x)^2)^(1/2)/sin(1/2*a+1/4*Pi+1/2*b*x)*Ell
ipticF(cos(1/2*a+1/4*Pi+1/2*b*x),2^(1/2))*sin(b*x+a)^(1/2)/b/c^2/(c*sin(b*x+a))^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2716, 2721, 2720} \[ \int \frac {1}{(c \sin (a+b x))^{5/2}} \, dx=\frac {2 \sqrt {\sin (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right ),2\right )}{3 b c^2 \sqrt {c \sin (a+b x)}}-\frac {2 \cos (a+b x)}{3 b c (c \sin (a+b x))^{3/2}} \]

[In]

Int[(c*Sin[a + b*x])^(-5/2),x]

[Out]

(-2*Cos[a + b*x])/(3*b*c*(c*Sin[a + b*x])^(3/2)) + (2*EllipticF[(a - Pi/2 + b*x)/2, 2]*Sqrt[Sin[a + b*x]])/(3*
b*c^2*Sqrt[c*Sin[a + b*x]])

Rule 2716

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
))), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \cos (a+b x)}{3 b c (c \sin (a+b x))^{3/2}}+\frac {\int \frac {1}{\sqrt {c \sin (a+b x)}} \, dx}{3 c^2} \\ & = -\frac {2 \cos (a+b x)}{3 b c (c \sin (a+b x))^{3/2}}+\frac {\sqrt {\sin (a+b x)} \int \frac {1}{\sqrt {\sin (a+b x)}} \, dx}{3 c^2 \sqrt {c \sin (a+b x)}} \\ & = -\frac {2 \cos (a+b x)}{3 b c (c \sin (a+b x))^{3/2}}+\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right ),2\right ) \sqrt {\sin (a+b x)}}{3 b c^2 \sqrt {c \sin (a+b x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.71 \[ \int \frac {1}{(c \sin (a+b x))^{5/2}} \, dx=-\frac {2 \left (\cos (a+b x)+\operatorname {EllipticF}\left (\frac {1}{4} (-2 a+\pi -2 b x),2\right ) \sin ^{\frac {3}{2}}(a+b x)\right )}{3 b c (c \sin (a+b x))^{3/2}} \]

[In]

Integrate[(c*Sin[a + b*x])^(-5/2),x]

[Out]

(-2*(Cos[a + b*x] + EllipticF[(-2*a + Pi - 2*b*x)/4, 2]*Sin[a + b*x]^(3/2)))/(3*b*c*(c*Sin[a + b*x])^(3/2))

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.36

method	result	size
default	\(-\frac {\sqrt {-\sin \left (b x +a \right )+1}\, \sqrt {2 \sin \left (b x +a \right )+2}\, \left (\sin ^{\frac {5}{2}}\left (b x +a \right )\right ) F\left (\sqrt {-\sin \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \left (\sin ^{3}\left (b x +a \right )\right )+2 \sin \left (b x +a \right )}{3 c^{2} \sin \left (b x +a \right )^{2} \cos \left (b x +a \right ) \sqrt {c \sin \left (b x +a \right )}\, b}\)	\(105\)

[In]

int(1/(c*sin(b*x+a))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/3/c^2*((-sin(b*x+a)+1)^(1/2)*(2*sin(b*x+a)+2)^(1/2)*sin(b*x+a)^(5/2)*EllipticF((-sin(b*x+a)+1)^(1/2),1/2*2^
(1/2))-2*sin(b*x+a)^3+2*sin(b*x+a))/sin(b*x+a)^2/cos(b*x+a)/(c*sin(b*x+a))^(1/2)/b

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.11 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.65 \[ \int \frac {1}{(c \sin (a+b x))^{5/2}} \, dx=\frac {{\left (\sqrt {2} \cos \left (b x + a\right )^{2} - \sqrt {2}\right )} \sqrt {-i \, c} {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + {\left (\sqrt {2} \cos \left (b x + a\right )^{2} - \sqrt {2}\right )} \sqrt {i \, c} {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) + 2 \, \sqrt {c \sin \left (b x + a\right )} \cos \left (b x + a\right )}{3 \, {\left (b c^{3} \cos \left (b x + a\right )^{2} - b c^{3}\right )}} \]

[In]

integrate(1/(c*sin(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

1/3*((sqrt(2)*cos(b*x + a)^2 - sqrt(2))*sqrt(-I*c)*weierstrassPInverse(4, 0, cos(b*x + a) + I*sin(b*x + a)) +
(sqrt(2)*cos(b*x + a)^2 - sqrt(2))*sqrt(I*c)*weierstrassPInverse(4, 0, cos(b*x + a) - I*sin(b*x + a)) + 2*sqrt
(c*sin(b*x + a))*cos(b*x + a))/(b*c^3*cos(b*x + a)^2 - b*c^3)

Sympy [F]

\[ \int \frac {1}{(c \sin (a+b x))^{5/2}} \, dx=\int \frac {1}{\left (c \sin {\left (a + b x \right )}\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(1/(c*sin(b*x+a))**(5/2),x)

[Out]

Integral((c*sin(a + b*x))**(-5/2), x)

Maxima [F]

\[ \int \frac {1}{(c \sin (a+b x))^{5/2}} \, dx=\int { \frac {1}{\left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate(1/(c*sin(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

integrate((c*sin(b*x + a))^(-5/2), x)

Giac [F]

\[ \int \frac {1}{(c \sin (a+b x))^{5/2}} \, dx=\int { \frac {1}{\left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate(1/(c*sin(b*x+a))^(5/2),x, algorithm="giac")

[Out]

integrate((c*sin(b*x + a))^(-5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(c \sin (a+b x))^{5/2}} \, dx=\int \frac {1}{{\left (c\,\sin \left (a+b\,x\right )\right )}^{5/2}} \,d x \]

[In]

int(1/(c*sin(a + b*x))^(5/2),x)

[Out]

int(1/(c*sin(a + b*x))^(5/2), x)